# The Method of Characteristics 1 Homogeneous transport equations

```The Method of Characteristics
Recall that the first order linear wave equation
ut + cux = 0,
u(x, 0) = f (x)
is constant in the direction (1, c) in the (t, x)-plane, and is therefore constant
on lines of the form x − ct = x0 . To determine the value of u at (x, t), we go
backward along these lines until we get to t = 0, and then determine the
value of u from the initial condition. The result is u(x, t) = u(x − ct, 0) =
f (x − ct).
There are many extensions to this simple idea. We begin by describing
the situation for linear and nearly linear equations.
1
Homogeneous transport equations
We can carry out this same idea for equations of the form
ut + c(x, t)ux = 0,
u(x, 0) = f (x),
−∞ < x < ∞.
(1)
Let X(T ) be any “trajectory” - think of it as a curve in the (x, t) plane, where
(X, T ) are supposed to be the (x, t) coordinates. How does u evolve as we
move along this trajectory?
d
u(X(T ), T ) = X 0 (T )ux (X(T ), T ) + ut (X(T ), T )
dT
by the chain rule. If we happen to pick X 0 (T ) = c(X(T ), T ), then
d
u(X(T ), T ) = c(X(T ), T )ux (X(T ), t) + ut (X(T ), T ) = 0
dT
by virtue of equation (1). Thus u is constant along ALL curves which are
solutions of the ODE X 0 (T ) = c(X, T ). To solve for u at some (x, t), we go
“backward” along this curve until we hit time zero, and since u is constant
along this curve, we find that the value of u is determined by the initial
condition. In other words, if X(t) = x, then u(x, t) = u(X(0), 0) = f (X(0)).
The curves X(T ) that solve the ODE
X 0 (T ) = c(X, T ),
X(t) = x,
(2)
are called characteristics. For the purpose of finding characteristics, (x, t) are
fixed constants, and it is X and T that vary along characteristics.
1
Example 1. Solve ut + xux = 0 with initial condition u(x, 0) = cos(x).
Solution. A characteristic curve ending at (x, t) will solve
X 0 (T ) = X(T ),
X(t) = x,
whose solution is X(T ) = x exp(T − t). Since u is constant along the characteristic,
u(x, t) = u(X(0), 0) = cos(xe−t ).
Example 2. We want to solve
yux = xuy ,
u(0, y) = 2y 2 for y > 0
(3)
Solution. This equation can be written in the form (1) as
ux −
x
uy = 0,
y
treating x like the time variable. Let Y (X) denote characteristic curves,
which is a solution to
X
= 0.
Y 0 (X) −
Y
Separating variables Y dY = −XdX leads to X 2 + Y 2 = C; in other words,
characteristics are closed curves encircling the origin. If an implicitly defined characteristic curve passes through (x, y), it is described by X 2 +Y 2 =
x2 + y 2 . Since the solution is constant along this curve, setting X = 0 and
using the side condition in (3) gives
u(x, y) = u(X, Y ) = 2Y 2 = 2(x2 + y 2 ).
Notice that if a boundary condition were imposed on the entire y-axis,
then characteristic curves would intersect this boundary both at (0, y) and
(0, −y). Unless u(0, y) = u(0, −y), this problem would not have a solution.
1.1
Inhomogeneous transport equations
We can also solve equations of the form
ut + c(x, t)ux = g(u, x, t),
u(x, 0) = f (x),
−∞ < x < ∞.
(4)
The only difference between this and equation (1) is that u is not constant
along characteristics, but evolves according to
d
u(X(t), t) = g(u, X(t), t).
dt
2
(5)
In other words, if we let U (T ) = u(X(T ), T ) be the solution restricted to a
single characteristic, it solves an initial value problem, namely
U 0 (T ) = g(U, X(T ), T ),
U (0) = u(X(0), 0) = f (X(0)).
Thus, to find u at some point (x, t), we go backwards along the characteristic
that ends at x until time zero, then solve the ODE (5) forwards until T = t.
1.2
The method of characteristics for linear problems
We can summarize ideas above as an algorithm:
1. Find the characteristic terminating at (x, t): Solve X 0 (T ) = c(X, T )
with the “final” condition X(t) = x. Note that the solution for X(T )
will depend on x and t as parameters.
2. Determine the solution along a characteristic: Solve U 0 (T ) = g(U, X(T ), T )
subject to initial condition U (0) = U (X(0), 0). Again the solution depends on x and t as parameters.
3. Find the solution at the endpoint of the characteristic: The solution
of the PDE at (x, t) is simply u(x, t) = U (t).
Here are a couple examples of how this is used.
Example 1. Solve
ut + (x + t)ux = t,
u(x, 0) = f (x).
Solution. Characteristic curves solve the ODE
X 0 (T ) = X + T,
X(t) = x.
This equation has a particular solution, Xp = −T − 1; the general solution
is therefore X(T ) = CeT − T − 1. Using the condition X(t) = x, we find
that
X(T ) = eT −t (x + t + 1) − T − 1.
Now we need to find how u changes along the characteristic. We solve
U 0 (T ) = T,
U (0) = f (X(0)) = f (e−t (x + t + 1) − 1).
whose solution by direct integration is
1
U (T ) = f (e−t (x + t + 1) − 1) + T 2 .
2
3
Finally, the solution at (x, t) is simply the value at the endpoint of the characteristic
1
u(x, t) = U (t) = f (e−t (x + t + 1) − 1) + t2 .
2
Example 2. Solve the nonlinear problem
ut + 3ux = −u2 ,
u(x, 0) = f (x).
Solution. In this case, characteristics solve X 0 (T ) = 2 with X(t) = x, so
that X = 2(T − t) + x. Along each characteristic, the solution evolves as
U 0 (T ) = −U 2 (T ) with U (0) = f (X(0)) = f (−2t + x). This is nonlinear,
but we can solve it since it is just a separable ODE which can be written
dU/U 2 = −dT , so that integration gives 1/U = T + C. Using the initial
1
condition, one gets C = f (x−2t)
and
U (T ) =
1
T+
.
1
f (x−2t)
The final solution is obtained by setting u(x, t) = U (t).
Example 3. Suppose water flows over a landscape whose elevation is described by h(x). A simple model for surface water flow says that the flow
velocity is equal (in the right units) to −h0 (x). It follows that if u(x, t) is the
depth of water, then the flux of u is J = −h0 (x)u. In the absence of sources
u satisfies the conservation equation ut + (−h0 (x)u)x = 0, which can be
written in the form (4) as
ut − h0 (x)ux = h00 (x)u.
(6)
The term on the right accounts for the fact that water will accumulate in
valleys where h00 > 0, and is depleted from hills where h00 < 0.
Consider a simple model for a valley where h = x2 , and suppose that
the initial depth is localized as
(
1 |x| ≤ 1
u(x, 0) =
0 |x| > 1
Since equation (6) reads ut − 2xux = 2u, characteristics solve X 0 (T ) = −2X
together with the terminal condition X(t) = x. The solution of this problem
is
X(T ) = xe2(t−T ) .
4
The solution on characteristics now solves U 0 = 2U with initial condition
(
1 |X(0)| ≤ 1
U (0) =
0 |X(0)| > 1
Therefore U (T ) = e2T if |X(0)| = |xe2t | < 1, or zero otherwise. It follows
that when t = T ,
(
e2t |x| ≤ e−2t
u(x, t) =
0
|x| > e−2t
The depth of the fluid layer therefore increases
R exponentially, but its width
decreases exponentially in a way such that udx remains constant.
5
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